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3 years ago

How many ways are there to distribute n graded homeworks to n students such that none of them gets their own homework back?

1 answer:

8 0

Assume N students

Student 1 can get (n-1) papers
Student 2 can get (n-1) papers
Student 3 can get (n-1) papers
etc
Student N can get (n-1) papers

So for N students you can have N(n-1)

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The cylindrical water tank on a semitrailer has a length of 20 feet. The volume of the tank is equal to the product of pi, the s

Bogdan [553]

Answer:

V = 20πr²

Step-by-step explanation:

In this question, volume of the cylindrical tank has been defined as,

Volume V of the tank = Product of π, the square of the radius of the tank 'r' and the length of the tank 'h'.

Therefore, formula to calculate the formula will be,

V = πr²h

If the length of the cylindrical trailer is,

h = 20 feet

Then the volume of the trailer will be,

V = πr²(20)

V = 20πr²

7 0

3 years ago

Can anyone please help with this question?

vitfil [10]

The middle section is n(n+1)

The top row is n+2

The bottom row is 2n+1

So the rule for the total number of tiles will be the sum of the above rules...

s(n)=n(n+1)+n+2+2n+1

s(n)=n^2+n+n+2+2n+1

s(n)=n^2+4n+3

6 0

3 years ago

whats 9+10?idk i cant seem to figure it out heeeeeeeeeeeeeeeeeeeeeeeeeeeelp me pleeeeeeeeeeeeeeeese. it the most hard math probe

KonstantinChe [14]

You cant fool me! Its either 19 or 21

7 0

3 years ago

Read 2 more answers

Which equation represents the line that is perpendicular to and passes through (-40,20)?

emmasim [6.3K]

The equation of the perpendicular line to the given line is: y = -5/4x - 30.

<h3>What is the Equation of Perpendicular Lines?</h3>

The slope values of two perpendicular lines are negative reciprocal of each other.

Given that the line is perpendicular to y = 4/5x+23, the slope of y = 4/5x+23 is 4/5. Negative reciprocal of 4/5 is -5/4.

Therefore, the line that is perpendicular to it would have a slope (m) of -5/4.

Plug in m = -5/4 and (x, y) = (-40, 20) into y = mx + b to find b:

20 = -5/4(-40) + b

20 = 50 + b

20 - 50 = b

b = -30

Substitute m = -5/4 and b = -30 into y = mx + b:

y = -5/4x - 30

The equation of the perpendicular line is: y = -5/4x - 30.

Learn more about about equation of perpendicular lines on:

brainly.com/question/7098341

#SPJ1

8 0

1 year ago

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$

8 0

1 year ago